How many pairs of integers exist for which the difference between their sum and product is 42

To solve the problem of finding how many pairs of integers exist such that the difference between their sum and product is 42, we can set this up mathematically.

Let the two integers be $x$ and $y$. We are given the condition:

$x + y - xy = 42$

We can rearrange this equation:

$xy - x - y = -42$

Adding 1 to both sides gives:

$xy - x - y + 1 = -41$

This can be factored:

$(x - 1)(y - 1) = -41$

Now we need to determine the integer factor pairs of $-41$. The integer factors of $-41$ are:

1. $(-1, 41)$
2. $(1, -41)$
3. $(-41, 1)$
4. $(41, -1)$

For each pair of factors $(a, b)$, we substitute back to find $x$ and $y$:

1. From $(a, b) = (-1, 41)$:

   • $x - 1 = -1 \Rightarrow x = 0$
   • $y - 1 = 41 \Rightarrow y = 42$
   • Pair: $(0, 42)$

2. From $(a, b) = (1, -41)$:

   • $x - 1 = 1 \Rightarrow x = 2$
   • $y - 1 = -41 \Rightarrow y = -40$
   • Pair: $(2, -40)$

3. From $(a, b) = (-41, 1)$:

   • $x - 1 = -41 \Rightarrow x = -40$
   • $y - 1 = 1 \Rightarrow y = 2$
   • Pair: $(-40, 2)$

4. From $(a, b) = (41, -1)$:

   • $x - 1 = 41 \Rightarrow x = 42$
   • $y - 1 = -1 \Rightarrow y = 0$
   • Pair: $(42, 0)$

Final Count

The valid pairs of integers $(x, y)$ are:

1. $(0, 42)$
2. $(2, -40)$
3. $(-40, 2)$
4. $(42, 0)$

Thus, there are 4 pairs of integers that satisfy the given condition.

Final Answer

There are 4 pairs of integers $(x, y)$ such that the difference between their sum and product is 42.